How do you test the improper integral #int (3x)/(x+1)^4 dx# from #[0, oo)# and evaluate if possible?

Question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T21:54:30

#int_0^oo(3x)/(x+1)^4dx=1/2#

#int_0^oo(3x)/(x+1)^4dx#

First just treating without the bounds:

#int(3x)/(x+1)^4dx#

Let #u=x+1#. Thus #du=dx# and #x=u-1#:

#=int(3(u-1))/u^4du=int(3/u^3-3/u^4)du#

Rewriting and integrating:

#=int(3u^-3-3u^-4)du=(3u^-2)/(-2)-(3u^-3)/(-3)#

Since #u=x+1#:

#=(-3)/(2(x+1)^2)+1/(x+1)^3=(-3(x+1)+2)/(2(x+1)^3)=(-3x-1)/(2(x+1)^2)#

So, we have:

#int_0^oo(3x)/(x+1)^4dx=[(-3x-1)/(2(x+1)^2)]_0^oo#

In order to "evaluate" this at infinity, take the limit:

#=(lim_(xrarroo)(-3x-1)/(2(x+1)^2))-((-3(0)-1)/(2(0+1)^2))#

The degree of the numerator exceeds that of the denominator, so the limit is #0#.

#=0-((-1)/(2))#

#=1/2#